Four Incircles in Equilateral Triangle: What Is This About?
A Mathematical Droodle

This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at, download and install Java VM and enjoy the applet.

What if applet does not run?


|Activities| |Contact| |Front page| |Contents| |Geometry| |Eye opener| |Store|

Copyright © 1996-2017 Alexander Bogomolny

Four Incircles in Equilateral Triangle

Here is a sangaku - a Temple geometry - problem in an equilateral triangle similar to another one in a square. The similarity aside, the current one is much more difficult than the other.

This is the problem #2.1.7 from [Fukagawa and Pedoe]. The problem is from an existent 1886 tablet found in the Fukusima prefecture:

In an equilateral triangle ABC the lines AC'A', BA'B', CB'C' are drawn making equal angles with AB, BC, CA, respectively, forming the triangle A'B'C', and so that the radius of the incircle of triangle A'B'C' is equal to the radius of the incircle of triangle AA'B. Find A'B' in terms of AB.

This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at, download and install Java VM and enjoy the applet.

What if applet does not run?

The answer to the problem is

A'B' = AB·3 / (3 + 21).

Unhappily I must confess that the equations I am able to write appear unappealing. I just could not see any shortcuts that would lead to an elegant solution.

Having no solution, I do not know whether the following observation is relevant or not: lines AA', BB', CC' touch, by the construction three of the four incircles. When the four have equal radii, the lines parallel to the former but passing on the other side of the incircle of ΔA'B'C' seem to also touch three of the four circles.

A later addition: George Zettler kindly sent me the following solution (6/16/2008).

First up, let's establish some general rules about triangles containing one angle of 120° (let's call them 120-triangles). We will denote the sides of such a thing as {a, b, c} where "c" will denote the side opposite the 120° angle.

Claim 1: the radius of the incircle in a 120-triangle is 3(a + b - c)/2.

Proof: This goes much like the analogous calculation for a right triangle. We have the picture:

We remark that r, the radius in question, is 3z (this is an easy calculation, using the 120° angle, and the kite which contains it). But it is obvious that c = a + b - 2z and the formula follows at once.

Claim 2: c² = a² + b² + ab.

Proof: this is a variant of the Pythagorean Theorem (or, if you know trig, it is just a special case of the Law of Cosines). From the picture we have:

The right triangle on the left is clearly a 30-60-90 triangle, from which it follows that q = a/2, and that p = 3a/2. The claim follows at once from the Pythagorean Theorem applied to the big right triangle.

Now, we need to look at the temple problem. The 3 congruent triangles are clearly 120-triangles. The inner one is equilateral. It is easy to work out the radius of the incircle in an equilateral triangle of side length x, it is x/(23).

The rest is algebra! Granted, it isn't terribly pretty algebra. We divide the sides of the 120-triangles into x and y, with x the side of the equilateral triangle in the middle. We want to find x and y; we have one equation from the radii and another from the Pythagorean Theorem. Here's the picture:

Ok, first the radii. Combining the result of Claim 1 with the simple calculation inside an equilateral triangle, we get:

r = 3(2y + x - 1)/2 = x / (23)

from which

x = 6y + 3x - 3


y = 1/2 - x/3.

Now the Law of Cosines:

1 = y² + (x + y)² + (x + y)y.

Substituting in for y, we get the following quadratic equation for x:

1 = (1/2 - x/3)² + (2x/3 + 1/2)² + (2x/3 + 1/2)(1/2 - x/3).

This quadratic equation simplifies to

x²/3 + x/2 - 1/4 = 0

which we can solve to get

x = (21 - 3) / 4.

Note: There is another proof by Angela Drei and a third one by J. John Samuel.


  1. H. Fukagawa, D. Pedoe, Japanese Temple Geometry Problems, The Charles Babbage Research Center, Winnipeg, 1989

    Write to:

    Charles Babbage Research Center
    P.O. Box 272, St. Norbert Postal Station
    Winnipeg, MB
    Canada R3V 1L6


  1. Sangaku: Reflections on the Phenomenon
  2. Critique of My View and a Response
  3. 1 + 27 = 12 + 16 Sangaku
  4. 3-4-5 Triangle by a Kid
  5. 7 = 2 + 5 Sangaku
  6. A 49th Degree Challenge
  7. A Geometric Mean Sangaku
  8. A Hard but Important Sangaku
  9. A Restored Sangaku Problem
  10. A Sangaku: Two Unrelated Circles
  11. A Sangaku by a Teen
  12. A Sangaku Follow-Up on an Archimedes' Lemma
  13. A Sangaku with an Egyptian Attachment
  14. A Sangaku with Many Circles and Some
  15. A Sushi Morsel
  16. An Old Japanese Theorem
  17. Archimedes Twins in the Edo Period
  18. Arithmetic Mean Sangaku
  19. Bottema Shatters Japan's Seclusion
  20. Chain of Circles on a Chord
  21. Circles and Semicircles in Rectangle
  22. Circles in a Circular Segment
  23. Circles Lined on the Legs of a Right Triangle
  24. Equal Incircles Theorem
  25. Equilateral Triangle, Straight Line and Tangent Circles
  26. Equilateral Triangles and Incircles in a Square
  27. Five Incircles in a Square
  28. Four Hinged Squares
  29. Four Incircles in Equilateral Triangle
  30. Gion Shrine Problem
  31. Harmonic Mean Sangaku
  32. Heron's Problem
  33. In the Wasan Spirit
  34. Incenters in Cyclic Quadrilateral
  35. Japanese Art and Mathematics
  36. Malfatti's Problem
  37. Maximal Properties of the Pythagorean Relation
  38. Neuberg Sangaku
  39. Out of Pentagon Sangaku
  40. Peacock Tail Sangaku
  41. Pentagon Proportions Sangaku
  42. Proportions in Square
  43. Pythagoras and Vecten Break Japan's Isolation
  44. Radius of a Circle by Paper Folding
  45. Review of Sacred Mathematics
  46. Sangaku à la V. Thebault
  47. Sangaku and The Egyptian Triangle
  48. Sangaku in a Square
  49. Sangaku Iterations, Is it Wasan?
  50. Sangaku with 8 Circles
  51. Sangaku with Angle between a Tangent and a Chord
  52. Sangaku with Quadratic Optimization
  53. Sangaku with Three Mixtilinear Circles
  54. Sangaku with Versines
  55. Sangakus with a Mixtilinear Circle
  56. Sequences of Touching Circles
  57. Square and Circle in a Gothic Cupola
  58. Steiner's Sangaku
  59. Tangent Circles and an Isosceles Triangle
  60. The Squinting Eyes Theorem
  61. Three Incircles In a Right Triangle
  62. Three Squares and Two Ellipses
  63. Three Tangent Circles Sangaku
  64. Triangles, Squares and Areas from Temple Geometry
  65. Two Arbelos, Two Chains
  66. Two Circles in an Angle
  67. Two Sangaku with Equal Incircles
  68. Another Sangaku in Square
  69. Sangaku via Peru
  70. FJG Capitan's Sangaku

Related material

  • Equilateral Triangles on Sides of a Quadrilateral
  • Euler Line Cuts Off Equilateral Triangle
  • Problem in Equilateral Triangle
  • Problem in Equilateral Triangle II
  • Sum of Squares in Equilateral Triangle
  • Triangle Classification
  • Isoperimetric Property of Equilateral Triangles
  • Maximum Area Property of Equilateral Triangles
  • Angle Trisectors on Circumcircle
  • Equilateral Triangles On Sides of a Parallelogram
  • Pompeiu's Theorem
  • Circle of Apollonius in Equilateral Triangle
  • |Activities| |Contact| |Front page| |Contents| |Geometry| |Eye opener| |Store|

    Copyright © 1996-2017 Alexander Bogomolny


    Search by google: